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Let the topological space X be the so-called "long line" — which is an uncountable linearly ordered set containing all the countable ordinal numbers as well as a copy of the open unit interval (0,1) between each countable ordinal number and its successor. The topology of X is its order topology. Let the topological space Y be the 2-dimensional sphere-more precisely, let Y be the boundary of the closed unit ball in 3-dimensional Euclidean space. Let Z be the topological product of X and Y.

My questions are:

  1. Is Z 3-dimensional, and if not, what is its dimension?
  2. Does Z contain an uncountable collection of pairwise disjoint open subsets?

I mean by "dimension", the so-called "small inductive dimension". This is a well known function, which for any non-empty regular Hausdorff space (as argument) has a non-negative integer value (or is infinite). The function has the same value for any pair of homeomorphic spaces in its range of definition. This includes Z-which is regular and Hausdorff, since both X and Y are.

The motivation for my questions is this: The physical space of our universe is most simply and clearly pictured as a 3-dimensional Euclidean space E(3) extending to infinity in all directions-in other words, Newtonian space, before the complications of relativity theory arose. Now E(3) contains an uncountable infinity of points but can only contain a countable infinity of "disjoint discrete objects" such as stars, galaxies or even atoms.

What simply described topological space could serve as the space of a physical universe containing uncountably many "disjoint discrete objects"? Such a space should probably contain an uncountable set of pairwise disjoint open subsets. A non-separable Hilbert space would meet this condition but its dimension is infinite. Could there not exist a low dimensional space, closer to E(3), which also meets this condition? That is why I am asking these questions about the arc-wise connected space Z. I also wonder whether Z shares the following desirable property of Euclidean spaces-being homogeneous and isotropic. If p and q are distinct points of Z, does there always exist a homeomorphism of Z onto itself that takes p into q?

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    $\begingroup$ Could you divide your question into paragraphs? It's a bit hard to read as a single block. $\endgroup$ – Joonas Ilmavirta Oct 7 '14 at 21:10
  • $\begingroup$ For your third question, $Z$ is a manifold and any manifold is homogeneous. $\endgroup$ – Mathieu Baillif Oct 8 '14 at 7:24
  • $\begingroup$ I would like to break up that whole pile of verbosity into $\endgroup$ – Garabed Gulbenkian Oct 9 '14 at 20:13
  • $\begingroup$ I would like to break up that whole pile of verbosity into separate sections or paragraphs but do not know how to do this. I tried several times but each attempt caused me to lose everything that I had written previously. $\endgroup$ – Garabed Gulbenkian Oct 9 '14 at 20:25
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  1. Dimension is a local property (there is a possibility that I am mistaken, please confirm) and the long line is locally homeomorphic to the real line, so $X$ has dimension 1. By the same argument $Z$ has dimension 3.
  2. Yes, and the open sets can even have pairwise disjoint closures. (More precisely, the closure of each open set will have a neighborhood that does not meet the closure of any other set in the collection.) It suffices to find an uncountable collection of open sets on $X$ with pairwise disjoint closures, since their products with $Y$ will give a similar collection for $Z$. Recall that $X=\omega_1\times[0,1)$. Let $A=\{\alpha+1\in\omega_1;\alpha\text{ is a limit ordinal}\}$. Now the open sets $\{\alpha\}\times(0,1)\subset X$, $\alpha\in A$, have disjoint closures and $A$ is uncountable.
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  • $\begingroup$ Thanks alot for your responses and for supplying paragraphs in my text. I now recall that there exist topological manifolds which are locally homeomorphic to 3-dimensional Euclidean space and which may (possibly) contain an uncountable collection of pairwise disjoint open subsets. If so, one of these may better represent the type of space that I am seeking, than Z does. I do not know if such manifolds are homogeneous or metrizable. $\endgroup$ – Garabed Gulbenkian Oct 10 '14 at 18:36
  • $\begingroup$ @GarabedGulbenkian, you are welcome. The existence of such manifolds depends on one's definition of a manifold. Some people require some kind of countability properties which exclude the long line, for example. You will certainly lose some of the nice Euclidean properties with such exotic spaces, but depending on your goal and tools that might not matter. $\endgroup$ – Joonas Ilmavirta Oct 10 '14 at 20:53
  • $\begingroup$ Notice that there are other non-metrizable $3$-manifolds which have the property of having pairwise disjoint open subsets, and even better: a discrete such collection. An example is $P\times\mathbb{R}$, where $P$ is any variant of the Prüfer surface (look for it in Wikipedia, for instance). And, again, any manifold is homogeneous. $\endgroup$ – Mathieu Baillif Oct 13 '14 at 18:37
  • $\begingroup$ I meant an uncountable discrete collection of open subsets, of course. $\endgroup$ – Mathieu Baillif Oct 13 '14 at 19:18

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